Question

An electron, helium ion (\( \text{He}^{++} \)) and proton having the same kinetic energy. The relation between their respective de-Broglie wavelengths \( \lambda_e, \lambda_{\text{He}^{++}} \) and \( \lambda_p \) is

• A. \( \lambda_e>\lambda_p>\lambda_{\text{He}^{++}} \)
• B. \( \lambda_e>\lambda_{\text{He}^{++}}>\lambda_p \)
• C. \( \lambda_e<\lambda_p<\lambda_{\text{He}^{++}} \)
• D. \( \lambda_e<\lambda_{\text{He}^{++}} = \lambda_p \)

Hint:

For same KE: wavelength depends only on mass $\longrightarrow$ lighter particle = larger wavelength.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The de-Broglie wavelength of a particle is related to its momentum. If multiple particles have the same kinetic energy, their wavelengths will depend inversely on the square root of their masses.
Step 2: Key Formula or Approach:
De-Broglie wavelength: \( \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \).
If kinetic energy (\( K \)) is constant, then \( \lambda \propto \frac{1}{\sqrt{m}} \).
Step 3: Detailed Explanation:
We compare the masses of the three particles:
1. Mass of electron (\( m_e \)) is approximately \( 9.1 \times 10^{-31} \text{ kg} \).
2. Mass of proton (\( m_p \)) is approximately \( 1.67 \times 10^{-27} \text{ kg} \).
3. Mass of helium ion (\( He^{++} \), which is an alpha particle) is approximately \( 4 \times m_p \).
So the order of masses is: \( m_e<m_p<m_{He^{++}} \).
Since \( \lambda \propto \frac{1}{\sqrt{m}} \), the particle with the smallest mass will have the largest wavelength.
Wavelength order: \( \lambda_e>\lambda_p>\lambda_{He^{++}} \).
Step 4: Final Answer:
The correct relation is \(\lambda_e>\lambda_p>\lambda_{He^{++}}\).